Coordinate Geometry

New member

There are two Points A(-1,0) and B(0,3).
(a) Find the gradient of AB. Answer: This is 3-0/0-(-1) = 3
(b) Find the equation of AB. Answer: y=3x+c, Therefore c = 3, and the equation is y=3x + 3
(c) If length of AB is √h, what is value of h. Answer:AB^2 =√(-1-0)^2 + (0-3)^2, so h=10
(d) If the point (-5,k) lies on the line produced, fine value of k. Answer, y=3(-5)+3 = -12. Therefore k is (-5, -12)
(e) If line y=x+1 is the line of symmetry of the triangle ABC, find the co-ordinates of C

Elite Member

There are two Points A(-1,0) and B(0,3).
(a) Find the gradient of AB. Answer: This is 3-0/0-(-1) = 3
(b) Find the equation of AB. Answer: y=3x+c, Therefore c = 3, and the equation is y=3x + 3
(c) If length of AB is √h, what is value of h. Answer:AB^2 =√(-1-0)^2 + (0-3)^2, so h=10
(d) If the point (-5,k) lies on the line produced, fine value of k. Answer, y=3(-5)+3 = -12. Therefore k is (-5, -12)
(e) If line y=x+1 is the line of symmetry of the triangle ABC, find the co-ordinates of C

Note that \(\displaystyle A\) is on the line \(\displaystyle y=x+1\) so \(\displaystyle C\) must be symmetric to \(\displaystyle B\) with respect to that line.
I suggest you draw a picture (no graph paper if you have some). If you do not know about lines of symmetry look here.

Elite Member

The problem definitely stated that \(\displaystyle y=x+1\) is an axis of symmetry.
As such, because \(\displaystyle A\) is on that line then the line must be the bisector if \(\displaystyle \angle BAC\).
The line \(\displaystyle y=-x+3\) will contain both \(\displaystyle B~\&~C\) and is perpendicular to \(\displaystyle y=x+1\)
Moreover, I think that the author wanted the point \(\displaystyle C\) to be the perpendicular distance from the line as is \(\displaystyle B\).

Yes, writing the distance formula and k=h+1 was my start, until I realized that midpoint formulas would not involve squared terms when solving the resulting system of two equations.

I also considered shifting the origin to (-1,0) and then viewing y=x+1 as y=x shifted. That way, AC would be the inverse of AB (reflected across the new y=x), so a run of 1 and a rise of 3 takes us from A to B and a run of 3 and a rise of 1 goes from A to C. (A better explanation of that would have been a lot more typing, but I'm still feeling wiped out from a trip to southern California.)